高中所有数学公式推导过程

2025-03-15 03:38:59
推荐回答(1个)
回答1:

对数的性质及推导
用^表示乘方,用log(a)(b)表示以a为底,b的对数
*表示乘号,/表示除号
定义式:
若a^n=b(a>0且a≠1)
则n=log(a)(b)
基本性质:
1.a^(log(a)(b))=b
2.log(a)(MN)=log(a)(M)+log(a)(N);
3.log(a)(M/N)=log(a)(M)-log(a)(N);
4.log(a)(M^n)=nlog(a)(M)
推导
1.这个就不用推了吧,直接由定义式可得(把定义式中的[n=log(a)(b)]带入a^n=b)
2.
MN=M*N
由基本性质1(换掉M和N)
a^[log(a)(MN)]
=
a^[log(a)(M)]
*
a^[log(a)(N)]
由指数的性质
a^[log(a)(MN)]
=
a^{[log(a)(M)]
+
[log(a)(N)]}
又因为指数函数是单调函数,所以
log(a)(MN)
=
log(a)(M)
+
log(a)(N)
3.与2类似处理
MN=M/N
由基本性质1(换掉M和N)
a^[log(a)(M/N)]
=
a^[log(a)(M)]
/
a^[log(a)(N)]
由指数的性质
a^[log(a)(M/N)]
=
a^{[log(a)(M)]
-
[log(a)(N)]}
又因为指数函数是单调函数,所以
log(a)(M/N)
=
log(a)(M)
-
log(a)(N)
4.与2类似处理
M^n=M^n
由基本性质1(换掉M)
a^[log(a)(M^n)]
=
{a^[log(a)(M)]}^n
由指数的性质
a^[log(a)(M^n)]
=
a^{[log(a)(M)]*n}
又因为指数函数是单调函数,所以
log(a)(M^n)=nlog(a)(M)
其他性质:
性质一:换底公式
log(a)(N)=log(b)(N)
/
log(b)(a)
推导如下
N
=
a^[log(a)(N)]
a
=
b^[log(b)(a)]
综合两式可得
N
=
{b^[log(b)(a)]}^[log(a)(N)]
=
b^{[log(a)(N)]*[log(b)(a)]}
又因为N=b^[log(b)(N)]
所以
b^[log(b)(N)]
=
b^{[log(a)(N)]*[log(b)(a)]}
所以
log(b)(N)
=
[log(a)(N)]*[log(b)(a)]
{这步不明白或有疑问看上面的}
所以log(a)(N)=log(b)(N)
/
log(b)(a)
性质二:(不知道什么名字)
log(a^n)(b^m)=m/n*[log(a)(b)]
推导如下
由换底公式[lnx是log(e)(x),e称作自然对数的底]
log(a^n)(b^m)=ln(a^n)
/
ln(b^n)
由基本性质4可得
log(a^n)(b^m)
=
[n*ln(a)]
/
[m*ln(b)]
=
(m/n)*{[ln(a)]
/
[ln(b)]}
再由换底公式
log(a^n)(b^m)=m/n*[log(a)(b)]
--------------------------------------------(性质及推导


公式三:
log(a)(b)=1/log(b)(a)
证明如下:
由换底公式
log(a)(b)=log(b)(b)/log(b)(a)
----取以b为底的对数,log(b)(b)=1
=1/log(b)(a)
还可变形得:
log(a)(b)*log(b)(a)=1
三角函数的和差化积公式
sinα+sinβ=2sin(α+β)/2·cos(α-β)/2
sinα-sinβ=2cos(α+β)/2·sin(α-β)/2
cosα+cosβ=2cos(α+β)/2·cos(α-β)/2
cosα-cosβ=-2sin(α+β)/2·sin(α-β)/2
三角函数的积化和差公式
sinα
·cosβ=1/2
[sin(α+β)+sin(α-β)]
cosα
·sinβ=1/2
[sin(α+β)-sin(α-β)]
cosα
·cosβ=1/2
[cos(α+β)+cos(α-β)]
sinα
·sinβ=-1/2
[cos(α+β)-cos(α-β)]

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